equation(1) h = 3t^2 a) find h when t=5 b)find t when h=108

for part AYou would but the t2 into brackets , then substitute t=5 to get h=3(52) using BIDMAS do the 52 = 25 so that h=3(25) so you end up with h=75
for part B first you would rearrange the equation to make t the subject (t on its own on one side) h/3=t2 then square root h/3sqrt{h/3}=t Then finally you substitute In h=108 sqrt{108/3}=tsqrt(3)=t

DN

Related Maths GCSE answers

All answers ▸

What is 2 1/5 + 1 3/4


What is completing the square?


Solve the equations x-y=1 and 5x-3y=13


Expand and simplify (x – 9)(x + 2)